Вопрос:

Решите неравенство: (x^2-x-12)/(x^2+x-2)>=0.

Ответ:

\[\frac{x^{2} - x - 12}{x^{2} + x - 2} \geq 0\]

\[x^{2} - x - 12 =\]

\[= x^{2} - 4x + 3x - 12 =\]

\[= x(x - 4) + 3(x - 4) =\]

\[= (x - 4)(x + 3)\]

\[x^{2} + x - 2 = x^{2} + 2x - x - 2 =\]

\[= x(x + 2) - (x + 2) =\]

\[= (x + 2)(x - 1)\]

\[\frac{(x + 3)(x - 4)}{(x + 2)(x - 1)} \geq 0\]

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